Question

The shortest distance between the lines $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}$ and $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}$ is

• A. $7 \sqrt{3}$
• B. $6 \sqrt{3}$
• C. $4 \sqrt{3}$
• D. $5 \sqrt{3}$

Answer 1

Answer: (B)

Shortest distance between two lines
$\frac{x-x_1}{a_1}=\frac{y-y_1}{a_2}=\frac{z-z_1}{a_3} \&$
$\frac{x-x_2}{b_1}=\frac{y-y_2}{b_2}=\frac{z-z_2}{b_3}$ is given as
$\frac{\begin{vmatrix}x_1-x_2 & y_1-y_2 & z_1-z_2 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{vmatrix}}{\sqrt{\left(a_1 b_3-a_3 b_2\right)^2+\left(a_1 b_3-a_3 b_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}$
$\frac{\begin{vmatrix}5-(3) & 2-(-5) & 4-1 \\ 1 & 2 & -3 \\ 1 & 4 & -5\end{vmatrix}}{\sqrt{(-10+12)^2+(-5+3)^2+(4-2)^2}}$
$\frac{\begin{vmatrix}8 & 7 & 3 \\ 1 & 2 & -3 \\ 1 & 4 & -5\end{vmatrix}}{\sqrt{(2)^2+(2)^2+(2)^2}}$
$ =\frac{|8(-10+12)-7(-5+3)+3(4-2)|}{\sqrt{4+4+4}} $
$ =\frac{16+14+6}{\sqrt{12}}=\frac{36}{\sqrt{12}}=\frac{36}{2 \sqrt{3}} $
$ =\frac{18}{\sqrt{3}}=6 \sqrt{3}$